The accompanying data resulted from an experiment in which weld diameter and shear strength (in pounds) were determined for five different spot welds on steel.Below are the data collected and the regression equation.Diameter Strength200.1 813.7210.1 785.3220.1 960.4230.1 1118.0240.0 1076.2Strength = -941.6992 + 8.5988*Diametera)The predicted y-hat value for a diameter of 201 is 864. Interpret this predicted value.b)what is the predicted strength of a weld with a diameter of 51?

Respuesta :

The equation of the least-squares line when y is expressed in kilograms is y = 2114.20 + 18.96x.

What is the linear system?

A Linear system is a system in which the degree of the variable in the equation is one. It may contain one, two, or more two variables.

The accompanying data resulted from an experiment in which weld diameter x and shear strength y (in pounds) were determined for five different spot welds on steel. A scatterplot shows a pronounced linear pattern.

Given the least square regression model equation:

y hat = -959.00 + 8.60x

weld diameter x ; shear strength y (in pounds)

1 lb = 0.4536 kg

To express shear strength y in kilogram:

y is multiplied by 0.4536

(y hat × 0.4536) = -959.00 + 8.60x

Divide both sides by 0.4536

(y hat × 0.4536) / 0.4536= (-959.00 + 8.60x) / 0.4536

y hat = 2114.1975 + 18.959435x

y hat = 2114.20 + 18.96x

The equation of the least-squares line when y is expressed in kilograms is y = 2114.20 + 18.96x.

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The accompanying data resulted from an experiment in which weld diameter x and shear strength y (in pounds) were determined for five different spot welds on steel. A scatterplot shows a pronounced linear pattern. The least-squares line is y hat = -959.00 + 8.60x. Because 1 lb = 0.4536 kg, strength observations can be re-expressed in kilograms through multiplication by this conversion factor: new y = 0.4536(old y). What is the equation of the least-squares line when y is expressed in kilograms? (Give the answer to two decimal places.)